Problem 7 - University Calculus: Early Transcendentals

Question

Use implicit differentiation to find \(d y / d x\). $$y^{2}=\frac{x-1}{x+1}$$

Step-by-Step Solution

Answer

\(dy/dx = 1/(y(x+1)^2)\).

Step 1: Differentiate Both Sides Implicitly

We start by differentiating both sides of the equation with respect to \(x\). The equation is \(y^2 = \frac{x-1}{x+1}\). The left side \(y^2\) is differentiated using the power rule and chain rule as \(2y \frac{dy}{dx}\).

Step 2: Differentiate the Right Side Using Quotient Rule

The right side \(\frac{x-1}{x+1}\) is a quotient. We differentiate it using the Quotient Rule:\[\frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{(x+1) \cdot 1 - (x-1) \cdot 1}{(x+1)^2} = \frac{(x+1)-(x-1)}{(x+1)^2} = \frac{2}{(x+1)^2}\]

Step 3: Set the Derivatives Equal

Combine both differentiated results: \[2y \frac{dy}{dx} = \frac{2}{(x+1)^2}\].

Step 4: Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) by dividing both sides by \(2y\):\[\frac{dy}{dx} = \frac{2}{2y(x+1)^2}\]. Simplifying gives:\[\frac{dy}{dx} = \frac{1}{y(x+1)^2}\].

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus used when differentiating functions that are expressed as a quotient, that is, a fraction with one function in the numerator and another in the denominator. In our original problem, we encounter the fraction \( \frac{x-1}{x+1} \). To differentiate it correctly, we apply the quotient rule.
The general formula for the quotient rule is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

Here, \( u = x-1 \) and \( v = x+1 \).
We differentiate each separately:

• The derivative of \( u = x-1 \) is \( 1 \).
• The derivative of \( v = x+1 \) is also \( 1 \).

Applying the quotient rule, we find:

• \( \frac{(x+1) \cdot 1 - (x-1) \cdot 1}{(x+1)^2} = \frac{(x+1)-(x-1)}{(x+1)^2} \)
• This simplifies to \( \frac{2}{(x+1)^2} \).

The quotient rule allows us to handle derivatives of fractions effectively, maintaining the integrity of each function within the expression.

Chain Rule

The chain rule is essential when differentiating composite functions, where one function is nested inside another. In the context of implicit differentiation, we often need the chain rule to handle variables that are not isolated.
Take the left side in our equation, \( y^2 \), which involves an inner function \( y \) squared. When differentiating this implicitly, we treat \( y \) as a function of \( x \), noting that its derivative will also involve \( \frac{dy}{dx} \).
For \( y^2 \), we apply both the power rule and the chain rule:

• Start with the power rule: the derivative of \( y^n \) is \( n\cdot y^{n-1} \).
• Incorporate the chain rule: multiply by \( \frac{dy}{dx} \) as \( y \) is a function of \( x \).

Combining these rules, the derivative of \( y^2 \) with respect to \( x \) becomes:

• \( 2y \frac{dy}{dx} \).

The chain rule is crucial in expressing derivatives in terms of \( x \) and \( y \) when implicating variables together, allowing us to connect their rates of change.

Power Rule

The power rule is a direct and straightforward tool for taking the derivative of polynomial functions. It states that if you have a function of the form \( x^n \), its derivative is \( n \cdot x^{n-1} \). This rule is part of our differential tactics when dealing with implicit differentiation.
In the exercise, we use the power rule in conjunction with the chain rule to differentiate \( y^2 \), viewing \( y \) as an implicit function of \( x \). By the power rule alone, differentiating \( y^2 \) gives:

• \( 2y \).

Then, applying the chain rule requires multiplying by \( \frac{dy}{dx} \) too:

• \( 2y \frac{dy}{dx} \).

The power rule provides simplicity in breaking down higher-order terms, especially when combined with other rules.
It is also implicitly embedded in the quotient rule when each component of the fraction is differentiated. This rule gives a concise means to tackle polynomial expressions effectively.